Solve by completing the square: x^2+15=8x

x^2+15=8x
x^2-8x=-15
x^2  - 2 (x)(4) =-15
adding 16 on both sides
x^2  - 2 (x)(4) +16=-15+16
x^2  - 2 (x)(4) + 4^2=  1

(x-4)^2=1

taking square root on bothsides   x-4 =  +-1
so x-4=1         or x-4=-1
x-5           or x=3

Aso sorry if i’m wrong

The first one, i believe, would be correct. all the others have one thing in common, and that's that they have repeating numbers. in the second one, x and y have repeating 6's and 3's, and in the third, x and y have repeating 6's as well. in the last one, there are repeating 3's. the first one has no repeating numbers.